设数列{bn}的前n项和为sn,且bn=2-2sn;数列{an}为等差数列,且a5=14,a7=20求
2个回答

1 = 2-2*b1

3b1 = 2

b1 = 2/3

bn - bn-1 = (2-2sn) - (2-2sn-1) = -2(sn- sn-1) = -2bn

3bn = bn-1

bn = 1/3 * bn-1

{bn}是等比数列

{bn} = { 2/3 * (1/3)^(n-1) } = { 2 * (1/3)^n }

a7 - a5 = 2d

d = (20 - 14 )/2 = 3

a1 = a5 - 4d = 14 - 4*3 = 2

{an} = { 2+ 3(n-1)} = { 3n -1 }

cn = an * bn = (2* (1/3)^n )*(3n-1)) = (6n-2) (1/3)^n

Tn = c1 + c2 +c3 + ...+ cn

Tn - 1/3 Tn =( c1 + c2 +c3 + ...+ cn) - 1/3(c1 + c2 +c3 + ...+ cn)

= c1 + (c2 - 1/3 c1) + (c3- 1/3 c2) + ...+ (cn -1/3 cn-1) - cn

= 4/3 + 6 * (1/3)^2 + 6 *(1/3)^3 + ...+ 6 * (1/3)^n - (6n-2)*(1/3)^n

= 4/3 + 6*( (1/3)^2 - (1/3)^(n+1)) / (1-1/3) ) -(6n-2)*(1/3)^n

= 4/3 + ( 1- (1/3)^(n-1) ) - (6n-2)*(1/3)^n

= 7/3 - (1/3)^(n-1) - (6n-2)*(1/3)^n

= 7/3 - 3*(1/3)^n -6n *(1/3)^n + 2 *(1/3)^n

= 7/3 - (1/3)^n - 6n *(1/3)^n

< 7/3

所以 2/3 *Tn < 7/3

Tn < 7/3 * 3/2 = 7/2